2 Answers
2

One way to show convergence of the first series, is to first note that
$$ \frac{4}{(n+1)(n+2)} < \frac{4}{(n+1)^2}$$ for all $n$. Then write
$$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)^2} = 4\sum_{n=2}^{\infty} \frac{1}{n^2}. $$ Presumably, you already know that the latter series converges, and so you can apply the comparison test, to show that
$$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)(n+2)} $$ converges. The other one can be treated in a similar way.

To find the sum, write
$$ \frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2},$$ and solve for $A$ and $B$ (this is called a partial fractions decomposition). Then you should be able to determine the $n$'th partial sum of this series explicitly, and easily evaluate the sum (these types of series will be called telescopic in your textbook, for obvious reasons). The other one can be treated in a similar way, but the computations will be a bit more complicated.

Note: Just realized that the first part of my answer is made redundant by the second, but I'll leave it - it can't hurt to know more than one way of accomplishing something.

For the second sum, take the terms $4\over (n+1)(n+3)$ and write them in the form
$$\tag{1}
{4\over (n+1)(n+3)}= {A\over n+1}+{B\over n+3}.
$$
We need to figure out what $A$ and $B$ are. Towards that end, multiply both sides of the above by $(n+1)(n+3)$. This gives
$$\tag{2}
4=A(n+3)+B(n+1).
$$
We're presuming this holds for any value of $n$. If we set $n=-3$, then equation $(2)$ becomes
$$
4=B(-2);
$$
which tells us $B=-2$.

Similarly, if we set $n=-1$ in equation $(2)$, we obtain $A=2$.

Thus, substituting these found values of $A$ and $B$ into equation $(1)$, we have
$$\tag{3}
{4\over (n+1)(n+3)}= {2\over n+1}-{2\over n+3}.
$$

The second sum in your question can then be written as
$$
\sum_{n=1}^\infty {4\over (n+1)(n+3)}=
\sum_{n=1}^\infty \biggl( {2\over n+1}-{2\over n+3}\biggr).
$$

Now, the infinite sum $\sum\limits_{n=1}^\infty a_n $ converges if and only if the sequence of partials sums $\{S_m\}$ defined by $S_m=\sum\limits_{n=1}^m a_n $ converges. In this case, we have
$$
\sum\limits_{n=1}^\infty a_n = \lim_{m\rightarrow\infty}\sum\limits_{n=1}^m a_n.
$$